Geometry problem of a triangle

Look at the figure. it is given a scalene triangle ABC. A square are drawn in each of its sides. If points P, Q and R are the centers of the square. Prove that: 1. RP perpendicular to AQ and 2. RP = AQ.

Easy Math Editor

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Comments

Here's a proof if you're familiar with complex numbers:

Let the complex numbers $a,b,c$ stand for the points on the triangle. Then $\large p=c+\frac{\sqrt{2}}{2}(a-c)e^{i\frac{-\pi}{4}}=c+\frac{\sqrt{2}}{2}(a-c)(\frac{1}{\sqrt{2}}(1-i))=c+\frac{1}{2}(a-c)(1-i)$

$\large p=\frac{1}{2}c(1+i)+\frac{1}{2}a(1-i)$

By analogy,

$\large q=\frac{1}{2}b(1+i)+\frac{1}{2}c(1-i)$

$\large r=\frac{1}{2}a(1+i)+\frac{1}{2}b(1-i)$

Giving,

$\large a-q= a-\frac{1}{2}b(1+i)-\frac{1}{2}c(1-i)$

$\large p-r= \frac{1}{2}c(1+i)+\frac{1}{2}a(1-i)-\frac{1}{2}a(1+i)-\frac{1}{2}b(1-i)$

$\large = \frac{1}{2}c(1+i)-ia-\frac{1}{2}b(1-i)$

$\large =-ia+\frac{1}{2}b(i-1)+ \frac{1}{2}c(1+i)$

$\large =-ia+\frac{1}{2}b(i-1)+ \frac{1}{2}c(i+1)$

$\large =-i(a-\frac{1}{2}b(1+i)- \frac{1}{2}c(1-i))$

$\large =e^{-i\frac{\pi}{2}}(a-q)$

QED

A L - 8 years ago

I was expecting proofs along the lines of complex numbers of vectors, which would be considered a standard exercise. There's a more basic approach, which uses similar ideas.

Hint: Let the square be labelled $ABDE$ . Consider triangles $ABQ$ and $EBC$ . Consider triangles $ARC$ and $AEC$ . Hence, we get fact 2. Show further that $EC$ makes a $45^\circ$ angle with both $AQ$ and $RP$ , which gives fact 1.

Calvin Lin Staff - 8 years ago

Thank u mr calvin for your hint. Just want to give a correction. Is it true that we should consider ARC and AEC, i think we should consider triangle ARP and AEC. Thank u. It was helping me a lot.

Falensius Nango - 8 years ago

thank u...

Falensius Nango - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$